dividing an offset circle into triangles First of all - I am sorry if it is the wrong forum or if this is a very trivial question. I am not a mathematician nor a trigonometry genius - and therefor I would ask a simple answer that someone like me could understand (and not just fancy formulas if possible)
Giving a circle with a known radius ($r$) and another circle with an offset of ($t$) 
I would like to fill the "gap" ($t$) with non-overlapping triangles with the closest possible angle to $45^\circ$.


*

*1 - How can I know how many triangles will enter the space ?

*2 - How can I calculate their exact angles (a) (b) ?

*3 - assuming I want an EXACT angle of (b) = $90^\circ$  (and not
approximation) - how can I know the number of triangles and also
calculate the "left-over" ??



UPDATE I:
as per comment : a visual example of wanted result.

 A: On this figure :

What you want is to have $c$ = $r_2 - d$. This leads you to the equation $\frac{r_2}{r_1} = \cos{\theta} + \sin{\theta}$. We assume here that $\theta > 0$. 
Use the trigonometry formulas to convert the cosine and sine as functions of $t = \tan{\frac{\theta}{2}}$, and this gives you t as the root of a degree $2$ polynomial :
$$
\begin{aligned}
  \frac{1+2t-t^2}{1+t^2} &= \frac{r_2}{r_1}
  \\
- \left(1+\frac{r_2}{r_1}\right)t^2
 +2t
+\left(1-\frac{r_2}{r_1}\right)
&= 0\\
\end{aligned}
$$
Solving this you get
$$
  \begin{aligned}
  \Delta &= 4\left(2-\frac{r_2^2}{r_1^2}\right)\\
  t &= \frac{
   1\pm \sqrt{2-\frac{r_2^2}{r_1^2}}
  }{
    \left(1+\frac{r_2}{r_1}\right)
  }
\end {aligned}
$$
and we assumed $\theta$ positive. 
Using $\arctan$, and rounding to find the best integer $k$ (since the number of samples is integral) such that $\theta = \frac{\pi}{k}$ should give you the angle $\theta$ to use as half of the sampling frequency for both circles.  Then both circles should be sampled using twice this angle frequency, but shifted with respect to one another with an angle $\theta$.
